# When can I pull out the identity operator?

I have a problem that requires figuring out the sum:

$\sum_{n>1} |\langle n10| \hat{z} |100 \rangle|^2 = \sum_{n>1} \langle 100|\hat{z}|n10 \rangle \langle n10|\hat{z}|100 \rangle$

where $|nlm\rangle$ are Hydrogen wave functions, and $\hat{z} = r\cos\theta$. For n=1, this term is zero; so then this sum is

$\sum_n ...|n10\rangle\langle n10| ...$

Is this the identity operator? Can I simply pull that out of this equation and say this is equal to $\langle 100|\hat{z}^2|100 \rangle$? I've checked these two values (my assumption which is just the one term, and the explicit sum to infinity) in Mathematica, and it seems like no. The expectation value of $\hat{z}^2$ with $|100\rangle$ is just $a_0^2$, whereas the sum (over 100 terms) seems to converge to about 0.717$a_0^2$. So it could be that a) I'm doing this right, and my Mathematica code is wrong, or b) my suspicions are confirmed that no, this is not the correct way to do this.

If this is the case, I'd also love some help actually computing this sum without doing hundreds of integrals.

Thanks! RS

• It is not the identity. May 3 '17 at 1:33
• It's not the identity because you aren't summing over a complete basis. May 3 '17 at 1:40
• I was thinking of your integrals... The problems with hydrogenoid radial integrals is that the argument of $R_{n\ell}$ actually depends on $(2r/na_0)$ and this makes the calculations of matrix elements between different $n$ quite non-trivial. I'm sure it's done somewhere, but I can't find exactly where. May 3 '17 at 1:55